Answer:
The answer to your question is None of your answers is correct, maybe the data are wrong.
Explanation:
Data
Concentration 1 = C1 = 1 M
Volume 2 = 5 ml
Concentration 2 = 0.05 M
Volume 1 = x
To solve this problem use the dilution formula
Concentration 1 x Volume 1 = Concentration 2 x Volume 2
Solve for Volume 1
Volume 1 = (Concentration 2 x Volume 2)/ Concentration 1
Substitution
Volume 1 = (0.05 x 5) / 1
Simplification
Volume 1 = 0.25/1
Result
Volume 1 = 0.25 ml
Answer: 11.2 moles of 
are produced when 5.60 mol of ethane is burned in an excess of oxygen.
Explanation:
The combustion of ethane is represented using balanced chemical equation:
As oxygen is preset in excess, ethane acts as the limiting reagent as it limits the formation of product.
According to stoichiometry :
2 moles of propane produces 4 moles of carbon dioxide
Thus 5.60 moles of propane will produce=
moles of carbon dioxide
Thus 11.2 moles of are produced when 5.60 mol of ethane is burned in an excess of oxygen.
Glucose is the major example of monosacchrides
succrose is an example of disaccharide
and both starch and cellulose are examples of polysaccharides
hope that helps :)
Answer:
The answer is the Third one
Hi! Glucose and oxygen are required for cellular respiration. As the law of conversation states, in a biochemical reaction, mass is conserved. For every glucose molecule, 6 oxygen molecules are used up and the end products, other than the energy dissipated by the reaction, are 6 water molecules and 6 carbon dioxide molecules. Brainliest?
